By definition (quasi-sober just drops uniqueness).

$T_0$ implies $x \ne y \implies \cl{\{x\}} \ne \cl{\{y\}}$. Thus, if $A = \cl{\{x\}}$, then this $x$ must be unique.

If there’s a basis of compact open sets, the ones intersecting $x$ form a local basis for $x$.

The open sets are trivially closed under finite intersections and form a basis. Since they are all compact, including $X$ itself, the space is spectral.

In order for $X \setminus \{p\}$ to be disconnected, it needs to have at least 2 points.

The space is a closed subspace of itself.

Finite implies countable.

$X$ is a countable union of finite open sets. $X$ hereditarily connected, so this basis is linearly ordered and it’s possible to enumerate them as $\{B_n\}$ where $n < m \implies B_n \subset B_m$. Therefore, $B_n \nearrow X$ and $X$ is a union of finite sets.

It’s almost… so not quite.

The homotopy cannot be empty.